ME 354 Tutorial 2B – Exergy – Control Volume Analysis
In the boiler of a typical power plant, H2O flows inside the tubes lining the walls of
the combustion chamber. Consider a case in which the H2O is brought from 0.8
MPa & 150C to 250C at essentially constant pressure while the combustion
gases passing over the tubes cool from 1100C to 500C at essentially a
constant pressure of 1 atm. The gases can be modeled as nitrogen (N 2). There is
no significant heat transfer from the combustion chamber to the surroundings.
H2O
T1 = 150°C
P1 = 0.8 MPa
.
1
mH20
.
3
H2O
T2 = 250°C
P2 = 0.8 MPa
.
mN2
N2
T3 =1100°C
mN2
4
N2
T3 =500°C
.
mH20
2
Find:
a) the mass flow rate of combustion gases per kg of H2O flowing inside the
tubes
b) the rate of exergy destruction per kg of H2O flow
c) the second law efficiency of the system
Step 1: Draw a diagram to represent the system
The first control volume encloses the H2O pipes as shown below
H2O
T1 = 150°C
P1 = 0.8 MPa
1
.
mH20
3
N2
T3 =1100°C
H2O
T2 = 250°C
P2 = 0.8 MPa
2
.
mN2
.
.
Q1
mN2
2
4
N2
T3 =500°C
.
mH20
The second control volume encloses all the N2 in the combustion chamber.
H2O
T1 = 150°C
P1 = 0.8 MPa
1
.
mH20
3
N2
T3 =1100°C
H2O
T2 = 250°C
P2 = 0.8 MPa
2
.
mN2
.
.
Q3
4
mN2
4
N2
T3 =500°C
.
mH20
1
Step 2: Prepare a property table
H2O
N2
State
1
2
3
T (°C)
150
250
1100
4
500
Property
P (MPa) h (kJ/kg) s (kJ/k*K)
0.8
0.8
P3
P3
Step 3: State your assumptions
Assumptions:
1) SSSF (steady state/steady flow)
2) Combustion gases modeled as N2 & Ideal Gas
3) No heat transfer from system to surroundings
4) ke, pe  0
5) Non-constant specific heats for N2
Step 4: Calculations (usually start by writing First and Second Laws)
Part a)
Writing the First Law for the control volume around the H2O tubes gives Eq1




dEcv
 m1 (e  Pv)1  m 2 (e  Pv) 2  Q W
dt
(Eq1)
Expanding the energy terms (e = u + ke + pe), realizing that there is no shaft
work
into
the
pipes,
and
applying
the
SSSF
assumption



dEcv
(
 0, m1  m 2  m H2O), Eq1 can be written as Eq2.
dt


0  m H 2O (u  ke  pe  Pv)1  (u  ke  pe  Pv) 2   Q1 2
(Eq2)
Using the assumption that ke and pe are approximately zero, and the fact the
property enthalpy, h, is defined as u + Pv, Eq2 can re-written as Eq3.


Q1 2  m H 2O h2  h1 
(Eq3)
Applying the same analysis to the N2 control volume (noting that the heat transfer
is now out of the system) yields Eq4.


- Q 34  m N 2 h4  h3 
(Eq4)
2
From examination of the control volume diagrams we can see that the heat
transfer from the N2 is equal to the heat transfer into the H2O. Eq5 expresses this
relation.


Q12  Q 34
(Eq5)
Substituting the relation for heat transfer in Eq3 & Eq4 into Eq5 gives


m H 2O h2  h1   m N 2 h3  h4 
(Eq6)

We need to find the ratio of the mass flow rate of combustion gases, m N 2 , to

mass flow rate of H2O, m H 2 O . Rearranging Eq6 we obtain Eq7.

mN 2

m H 2O

h2  h1
h3  h4
(Eq7)
We can use Table A-18 to determine the difference in enthalpy of N2 between
state 3 and 4. Note: Since we are looking at the difference in enthalpy we do not need to worry
about the reference point with which these enthalpies are determined. Also Note: Since there is
a 600 temperature difference between state 3 and 4, the constant specific heat assumption
could lead to considerable error. This is why we are not using the relation h4 - h3 = cp(T4-T3).
Table A-18 gives enthalpy on a kmol basis. To convert to a kg basis, we can use
Table A-1, which gives the molar mass of N2 as 28.02 kg/kmol.
From Table A-18, interpolating between T=1360K and T=1380K for T3=1373K
(1100C)
_
_
1373  1360
h 3  42227

 h 3  42674.2kJ / kmol
1380  1360 42915  42227
 h3 = (42674.2 kJ/kmol)/(28.02 kg/kmol) = 1523 kJ/kg
From Table A-18, interpolating between T=770K and T=780K for T 4=773K
(500C)
_
_
773  770
h 4  22772

 h 4  22865.9[kJ / kmol]
780  770 23085  22772
 h4 = (22865.9 kJ/kmol)/(28.02 kg/kmol) = 816 kJ/kg
To determine the enthalpy of H2O at state 1 and 2, we can use the steam tables.
At state 1, the temperature of the H2O is 150C @ 0.8MPa. Looking first in Table
A-5 at 0.8MPa we find that the corresponding saturated temperature is 170.43C.
Since T1 < 170.43C, the H2O at state 1 is subcooled i.e. a compressed liquid.
3
We can approximate the enthalpy at state 1 using the enthalpy of saturated liquid
at 150C (see p73 of text for explanation).
From Table A-4 @ 150C
 h1 = hf@150C =632.20 kJ/kg
At state 2, the temperature of the H2O is 250C @ 0.8MPa. Looking first in Table
A-5 at 0.8MPa we find that the corresponding saturated temperature is 170.43C.
Since T2 > 170.43C, the H2O at state 2 is superheated. We can use Table A-6
@ 0.8Mpa and 250C to determine the enthalpy of state 2.
From Table A-6 @ 250C, 0.8 MPa
 h2 = 2950.0 kJ/kg
We can now substitute these values into Eq7 to solve for the mass flow ratio.

mN2


m H 20
h2  h1 2950.0  632.20
= 3.28

h3  h4
1523  816
Answer a)
Part b)
H2O
T1 = 150°C
P1 = 0.8 MPa
1
.
mH20
3
N2
T3 =1100°C
H2O
T2 = 250°C
P2 = 0.8 MPa
2
.
.
mN2
mN2
4
N2
T3 =500°C
.
mH20

To find the rate of exergy destruction, X destroyed , we can perform an exergy
balance on the system as shown in Eq8.




X in  X out  X destroyed   X system
(Eq8)

Since we have assumed steady conditions,  X system  0 . With our choice of
boundary conditions, there is exergy transfer in and out of the system boundary
through mass flow only. Rewriting Eq8 to express the above information, we
arrive at Eq9.

 
 
X destroyed  m   m 

 IN 
 OUT
(Eq9)
4
The flow exergy terms in Eq9 can be expanded in terms of the individual flow
exergy terms of the system as shown in Eq10.




 

X destroyed  m H 2O  1  m N 2  3   m H 2O  2  m N 2  4 

 

(Eq10)
The question has asked for the exergy destruction per kg of H 2O flow. We can

obtain an expression for this by dividing Eq10 through by m H 2 O to obtain Eq11.


X destroyed

  1   2  
m H 2O
mN2

 3   4 
(Eq11)
m H 2O

We determined
mN 2
in part a), so the solution requires finding the differences in

m H 2O
the flow exergies. The difference in the flow exergy terms can be determined
from Eq12.
 1   2  h1  h2   T0 s1  s 2  

1 
v1
2
  v   g z
2
 2
2
1
 z2 
(Eq12)
Using the assumption that ke & pe  0, Eq12 reduces to Eq13.
1   2  h1  h2   T0 s1  s2 
(Eq13)
Similarly, applying the same steps for 3 – 4 we obtain Eq14.
 3   4  h3  h4   T0 s3  s4 
(Eq14)
We determined the enthalpies in part a) we now must similarly determine the
entropies.
From Table A-18, interpolating between T=1360K and T=1380K for T 3= 1373K
(1100C)
_
_
1373  1360
s 3  238.376
 kJ 

 s 3  238.702
1380  1360 238.878  238.376
 kmol  K 
 s3 = (238.702 kJ/kmol*K)/(28.02 kg/kmol) = 8.519 kJ/kg*K
5
From Table A-18, interpolating between T=770K and T=780K for T4= 773K
(500C)
_
_
773  770
s 4  219.709
 kJ 

 s 4  219.830
780  770 220.113  219.709
 kmol  K 
 s4 = (219.830 kJ/kmol)/(28.02 kg/kmol) = 7.845 kJ/kg*K
From Table A-4 @ 150C
 s1 = sf@150C =1.8418 kJ/kg*K
From Table A-6 @ 250C, 0.8 MPa
 s2 = 7.0384 kJ/kg*K
Substituting in these values in to Eq13 and Eq14.
  1  2  632.20  2950  2981.8418  7.0384 = -769.2 kJ/kg
  3   4  1523  816  2988.519  7.845 = 506.15 kJ/kg
Substituting the above into Eq11,

X destroyed

 769.2  3.28506.15 = 891 kJ/kgH2O
Answer b)
m H 2O
Part c)
The second law efficiency for this problem can be expressed as the increase in
exergy in the H2O divided by the reduction in exergy in the N2. This is expressed
in Eq15.

 II 
m H 2O  2   1 

m N 2  3   4 

769.2kJ / kg
= 0.463 or 46.3%
3.28506.15kJ / kg
Answer c)
Step 5: Concluding Statement & Remarks
The mass flow rate of combustion gases per kg of H2O flowing inside the tubes is
3.28, the rate of exergy destruction per kg of H2O is 891 kJ/kgH2O, and the
second law efficiency is 46.3%.
6